help me echelon matrix ?

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Nguyen Trong Nhan
Nguyen Trong Nhan on 5 Jan 2014
Answered: Roger Stafford on 6 Jan 2014
I have a matrix:
syms m
A = [1 2 3 4;2 -1 1 1;-1 4 3 2;1 2 0 m]
with m is the parameter I use rref command:
rref(A)
ans =
[ 1, 0, 0, 0]
[ 0, 1, 0, 0]
[ 0, 0, 1, 0]
[ 0, 0, 0, 1]
the parameter m was lost. How I can convert matrix A to echelon form that the parameter m is still kept. thanks you very much.

Accepted Answer

Matt J
Matt J on 5 Jan 2014
Edited: Matt J on 5 Jan 2014
I suspect, after experimenting with multiple m, that the echelon form of this particular matrix is independent of m, e.g.,
>> m=1; A = [1 2 3 4;2 -1 1 1;-1 4 3 2;1 2 0 m]; rref(A)
ans =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
>> m=10; A = [1 2 3 4;2 -1 1 1;-1 4 3 2;1 2 0 m]; rref(A)
ans =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1

More Answers (1)

Roger Stafford
Roger Stafford on 6 Jan 2014
It is in the nature of the reduced row echelon form for your square matrix to be the identity matrix. That is because no matter what value m has, the matrix is always non-singular.
Suppose we alter one number in A:
A = [1 2 3 4;2 -1 1 1;1 4 3 2;1 2 0 m]
Then rref(A) still gives the appearance of being independent of the value of m. However there is one and only one value, m = -3.2, which makes the matrix singular and in that case the bottom row of rref(A) will become all zeros, showing that it can be affected by the value of m.
You can read about this form and its properties at:
http://en.wikipedia.org/wiki/Row_echelon_form

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